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The angle of elevation of a jet fighter from a point A on the ground is $60 °$ . After a flight of 15 seconds, the angle of elevation changes to $30 °$ . If the jet is flying at a speed of $720 km / hr$ , find the constant height.

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2.59 km

3.59 km

4.59 km

5.59 km

answer is A.

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Detailed Solution

Given that the angle of elevation of a jet fighter from a point A on the ground is

60 °

.
We have to find the constant height.
The value for the trigonometric identities can be evaluated by using the right-angled triangle the value for

tan θ

can be expressed as

tan θ = opposite adjacent

Suppose, B and C be the two positions of the jet fighter and the constant height of the jet fighter be BE=CD=h.
Since the speed of jet fighter is

720 km / hr

Thus, the distance covered in 15 seconds is,

B C = 720 × 1000 × 15 3600 = 720 × 150 36 = 3000 m = 3 km

The required figure geometry is shown in the figure below,
Question Image

Then by using the figure geometry, in triangle BCA

tan θ = B E A E ⇒ tan 60 ° = h A E ⇒ 3 = h A E ⇒ A E = h 3

Similarly, in a triangle CDA

tan θ = C D A D ⇒ tan 30 ° = h A D ⇒ 13 = h A D ⇒ A D = h 3

A D = A E + E D ⇒ 3 h = h 3 + 3 ⇒ 2 h = 33 ⇒ h = 2.59 km

The constant height is 2.59 Km.
Therefore, the correct option is 1.

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